3.96.8 \(\int \frac {(4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} (-20-5\ 2^{5+8 x} x \log (2))) \log (\frac {1}{25} (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)))}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx\)

Optimal. Leaf size=24 \[ \log ^2\left (4+\left (-e^{2^{8 x}}+\frac {x}{5}\right )^2+\log (2)\right ) \]

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Rubi [A]  time = 0.54, antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 1, number of rules used = 2, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6684, 6686} \begin {gather*} \log ^2\left (\frac {1}{25} \left (x^2-10 e^{2^{8 x}} x+25 e^{2^{8 x+1}}+25 (4+\log (2))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 25*E^2^
(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])
,x]

[Out]

Log[(25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*(4 + Log[2]))/25]^2

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (\frac {1}{25} \left (25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 (4+\log (2))\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 7.01, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 2
5*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*L
og[2]),x]

[Out]

Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 2
5*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*L
og[2]), x]

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fricas [A]  time = 0.59, size = 29, normalized size = 1.21 \begin {gather*} \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2 \cdot 2^{8 \, x}\right )} + \log \relax (2) + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="fricas")

[Out]

log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2) + 4)^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (200 \cdot 2^{8 \, x} e^{\left (2 \cdot 2^{8 \, x}\right )} \log \relax (2) - 5 \, {\left (8 \cdot 2^{8 \, x} x \log \relax (2) + 1\right )} e^{\left (2^{8 \, x}\right )} + x\right )} \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2 \cdot 2^{8 \, x}\right )} + \log \relax (2) + 4\right )}{x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \relax (2) + 100}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="giac")

[Out]

integrate(4*(200*2^(8*x)*e^(2*2^(8*x))*log(2) - 5*(8*2^(8*x)*x*log(2) + 1)*e^(2^(8*x)) + x)*log(1/25*x^2 - 2/5
*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2) + 4)/(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100), x)

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maple [A]  time = 0.11, size = 26, normalized size = 1.08

method	result	size
risch	\(\ln \left ({\mathrm e}^{2 \,256^{x}}-\frac {2 x \,{\mathrm e}^{256^{x}}}{5}+\ln \relax (2)+\frac {x^{2}}{25}+4\right )^{2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))^2+(-160*x*ln(2)*exp(8*x*ln(2))-20)*exp(exp(8*x*ln(2)))+4*x)*
ln(exp(exp(8*x*ln(2)))^2-2/5*x*exp(exp(8*x*ln(2)))+ln(2)+1/25*x^2+4)/(25*exp(exp(8*x*ln(2)))^2-10*x*exp(exp(8*
x*ln(2)))+25*ln(2)+x^2+100),x,method=_RETURNVERBOSE)

[Out]

ln(exp(2*256^x)-2/5*x*exp(256^x)+ln(2)+1/25*x^2+4)^2

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maxima [B]  time = 0.55, size = 92, normalized size = 3.83 \begin {gather*} -\log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \relax (2) + 100\right )^{2} + 2 \, \log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \relax (2) + 100\right ) \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2^{8 \, x + 1}\right )} + \log \relax (2) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp(8*x*log(2))-20)*exp(exp(8*x*lo
g(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^
2-10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="maxima")

[Out]

-log(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100)^2 + 2*log(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2
^(8*x)) + 25*log(2) + 100)*log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2^(8*x + 1)) + log(2) + 4)

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mupad [B]  time = 8.25, size = 29, normalized size = 1.21 \begin {gather*} {\ln \left (\ln \relax (2)+{\mathrm {e}}^{2\,2^{8\,x}}-\frac {2\,x\,{\mathrm {e}}^{2^{8\,x}}}{5}+\frac {x^2}{25}+4\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(2) + exp(2*exp(8*x*log(2))) + x^2/25 - (2*x*exp(exp(8*x*log(2))))/5 + 4)*(4*x - exp(exp(8*x*log(2
)))*(160*x*exp(8*x*log(2))*log(2) + 20) + 800*exp(2*exp(8*x*log(2)))*exp(8*x*log(2))*log(2)))/(25*log(2) + 25*
exp(2*exp(8*x*log(2))) + x^2 - 10*x*exp(exp(8*x*log(2))) + 100),x)

[Out]

log(log(2) + exp(2*2^(8*x)) - (2*x*exp(2^(8*x)))/5 + x^2/25 + 4)^2

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sympy [A]  time = 2.10, size = 39, normalized size = 1.62 \begin {gather*} \log {\left (\frac {x^{2}}{25} - \frac {2 x e^{e^{8 x \log {\relax (2 )}}}}{5} + e^{2 e^{8 x \log {\relax (2 )}}} + \log {\relax (2 )} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))**2+(-160*x*ln(2)*exp(8*x*ln(2))-20)*exp(exp(8*x*ln(2))
)+4*x)*ln(exp(exp(8*x*ln(2)))**2-2/5*x*exp(exp(8*x*ln(2)))+ln(2)+1/25*x**2+4)/(25*exp(exp(8*x*ln(2)))**2-10*x*
exp(exp(8*x*ln(2)))+25*ln(2)+x**2+100),x)

[Out]

log(x**2/25 - 2*x*exp(exp(8*x*log(2)))/5 + exp(2*exp(8*x*log(2))) + log(2) + 4)**2

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